Find the 2x power derivative of function E,

(e^(2x))'=e^(2x)*(2x)'=e^(2x)*2=2e^(2x)
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What is the derivative of the SiNx power of y equal to x?

Using logarithmic derivative method
lny=sinx*lnx
Derivation of X at both ends of the above formula
y'/y=cosx*lnx+(sinx)/x
y'=y(cosx*lnx+(sinx)/x)=(x^(sinx))*(cosx*lnx+(sinx)/x)

It is proved that the x power of E is greater than 1 plus x, and X is not equal to zero

To prove that e ^ x > 1 + X is to prove that e ^ x-1-x > 0, and then the derivative of e ^ x-1-x is e ^ X-1. Then when x > 0, e ^ X-1 is always greater than 0, then e ^ x-1-x is increasing. When x = 0, e ^ x-1-x = 0, so e ^ x-1-x > 0 is always true when x > 0. At the same time, when x0 is always true for different x, e ^ x > 1 + X

F (x) = the third power of 2 / 3 x (x1), then the left derivative of F (x) exists and the right derivative does not exist at x = 1. How to find it,

F '(left, x = 1) = 2x ^ 2 = 2
F '(right, x = 1) = 2x = 2
The first derivative is continuous. It is also possible to take the limit
F '(right, x = 1) = LIM (DX - > 0, ((x + DX) ^ 2-x ^ 2) / DX) = 2

The derivative of (the x power of 2 times the x power of E times the x power of π)

y=2^xe^xπ^x=(2*e*π)^x
y’=(2*e*π)^xln(2*e*π)

Range options of function f (x) = 1 / 2 (SiNx + cosx) + 1 / 2 | SiNx cos | A. [1,1] B. [negative half root two, 1] C. [- 1 / 2,1 / 2] D. [- 1, half root two]

1) SiNx > cosx, then SiNx > - √ 2 / 2, f (x) = SiNx, -√ 2 / 22) SiNx < = cosx, then cosx > - √ 2 / 2, f (x) = cosx, -√ 2 / 2 < = f (x) < = 1
Therefore, the value range of F (x) is [- √ 2 / 2,1], select B

Find the 2x power derivative of function E,